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The math behind exploding dice rolls

Sun, 19 Sep 2021 00:00:00 GMT

Exploding dice is a mechanic that appears in various tabletop games. It works like this: a dice is rolled, if it lands on its maximum values, it “explodes” — the dice is then rolled again, adding the new value to the max value of the dice. If dice again lands on its maximum, the dice explodes a second time. This process continues and may lead to chains of explosions that push the roll total far beyond what is normally expected.

This mechanic appears in a number of tabletop games, with Savage Worlds likely being the most popular system within the RPG space that uses it. In this post, we dig into the infinite nature of exploding dice and explore the math and expected outcomes from this fun mechanic.

Exploding probabilities

Suppose we roll an $n$ sided exploding dice. The probability of rolling a non-exploded value, that is between $1$ and $n-1$ is no different than a normal dice roll. Each face below $n$ appears with equal probability $1/n$ . If instead the dice explodes — which happens with the remaining probability $1/n$ — we add $n$ to the outcome of a new exploding dice roll.

Let’s look at an example when we are rolling a 6-sided dice, so $n = 6$ . Any result between 1 and 5 appears with probability $1/6$ . How about the probability of rolling a 7? Rolling a 7 requires rolling a 6 on the first roll, triggering the explosion, and then rolling a 1 on the second:

P(X = 7) = P(X = 6)P(X=1) = \frac{1}{6^2}

Any value of the exploded dice roll can be written like this — as powers of $1/n$ . Suppose we want to know the probability of rolling a 25 on an exploding d6. Well that’s just rolling 6 four times in a row followed by a 1, so $p = 1/6^5$ . In general, for an $n$ sided dice, the probability of rolling any given value is:

\boxed{ P(X = x) = \begin{cases} \left(\frac{1}{n}\right)^{\lfloor{x/n}\rfloor + 1} & x \textrm{ not a multiple of }n \\ 0 & \textrm{otherwise} \end{cases} }

Where $\lfloor{x/n}\rfloor$ represents the rounded down result of dividing $x$ by $n$ . The probability of rolling a multiple of $n$ is 0, because every time a multiple of $n$ would show up, it triggers a dice explosion. The below graph shows the outcome probabilities when rolling an exploding d6.

Within a given number of explosions, each outcome is equally likely. As the number of explosions increases, the probability decays rapidly due to the rarity of rolling many 6s in a row. This particular graph cuts off at 18, but in reality it continues to infinity with ever decreasing probabilities.

The probability of at least $k$ explosions on an $n$ sided dice is $P(\textrm{at least k explosions}) = n^{-k}$ . Continuing with the d6 example, we have:

\begin{aligned} P(\textrm{at least 1 explosion on d6}) &\approx &17\% \\ P(\textrm{at least 2 explosions on d6}) &\approx &3\% \\ P(\textrm{at least 3 explosions on d6}) &\approx &0.5\% \\ P(\textrm{at least 4 explosions on d6}) &\approx &0.08\% \\ \end{aligned}

While one or two explosions is fairly common, more beyond that are quite rare, with a less than 1-in-1000 chance that you roll at least 4 explosions on a d6.

Explosive expectations

Another way to understand the effect of the exploding dice mechanic is to look at how it changes the average or expected outcome of a roll. In a previous post, we showed that the expectation of a normal $n$ sided dice roll is $(n + 1) / 2$ . It is clear that exploding dice would have a higher average, but by how much?

We’ll calculate this by appealing to something called the law of total expectation. It allows us to write the expectation of an exploding dice roll as follows:

E(X) = E(X | \textrm{explode}) P(\textrm{explode}) + E(X | \textrm{not explode}) P(\textrm{not explode})

By breaking up our expectation into two distinct cases — exploding and not exploding — we simplify our approach. First, the probability statements are are easily addressed by our previous work:

P(\textrm{explode}) = \frac{1}{n}

P(\textrm{not explode}) = \frac{n-1}{n}

Next, we consider the expected value of a roll with no explosion. Since each outcome below the maximum value is equally likely, this the same as the expected result of an $n-1$ sided roll. Plugging in to $(n + 1) / 2$ , we get:

E(X | \textrm{not explode}) = \frac{n}{2}

The last term, which is the expected value of an exploded dice is less clear, but we can get around it with a little trick. Recall that once a dice explodes, you can view it as a new exploding dice roll with $n$ added to the final result. We can express this as:

E(X | \textrm{explode}) = n + E(X)

While it may seem odd to write the original thing that we were trying to solve for — $E(X)$ — on the right hand side, it’s perfectly valid to do and will help us later. Now that we have expressions for all four terms in our original expectation equation, we can plug in to get:

E(X) = \left(n + E(X)\right) \frac{1}{n} + \frac{n}{2}\frac{n - 1}{n}

The above simplifies to:

E(X) = 1 + \frac{E(X)}{n} + \frac{n - 1}{2}

We take advantage of our trick by subtracting $E(X)/n$ from both sides to gather all the $E(X)$ terms on the right and solve to get:

\boxed{ E(X) = \frac{n}{n-1} \frac{n + 1}{2} }

Careful inspection of our final result shows that the expectation of a normal dice roll appears on in our result, $(n + 1)/2$ . A succinct way to summarize the effects of the exploding dice mechanic is that it increases the expected result by a factor of $n / (n - 1)$ .

Revisiting our $n=6$ case, we plug in to see that $E(X) = 4.2$ . A notable increase over the normal average of 3.5. The below graph summarizes the differences over the standard dice from d4 up to d20.

While the effect is of a similar magnitude across different types, it is most significant for low-faced dice. The factor of increase, $n / (n - 1)$ , quickly approaches 1 as $n$ increases, but with few sides it is relatively large. For a d4, the increase from an average roll of 2.5 to one of 3.33, represents a 33% increase. On the other hand, the d20’s 10.5 to 11.05 increase is a meager 5%. The graph below shows percent increase of the exploding mechanic across the standard dice types:

Despite explosions having a consistent increase the expected outcome, as a percent difference, the effect is attenuated for high-sided dice. Such a difference suggests some reasoning behind the prevalence of exploding dice in d6 systems, but less so in those with a greater d20 focus.

Regardless of those differences, exploding dice is a fun mechanic that leads to some interesting math and meaningfully alters outcomes in systems that employ it.

Rolling for stats in practice: how bad luck protection inflates ability scores

Sat, 17 Jul 2021 00:00:00 GMT

Rolling for stats in DnD 5e always carries the risk of ending up with a weak character. Despite recommendations to use the standard array, point buy, or any number of homebrew rules for generating stats, rolling rules-as-written nevertheless remains common. For these tables, poor rolls inevitably occur, and when that happens, many DMs implement bad luck protection by allowing bad stats to be rerolled.

In this post, we look at the mathematical consequences of rerolling your stat array. By setting a limit on how bad a player can roll before allowing a reroll — we call that limit the bad luck threshold — the underlying distribution of possible outcomes changes and biases ability scores higher than they would otherwise be.

Reviewing the standard array and rolling RAW

Before we dive into how rerolling changes the odds, let’s review how rolling rules-as-written compares to the standard array. As a reminder, the standard array is [8, 10, 12, 13, 14, 15]. Rolling for stats with 4d6 dropping the lowest generates a wide range of possible outcomes.

The chart below compares the two methods. The top-left shows the distribution of possible outcomes for the lowest value in your rolled array; typical values are somewhere between 6 and 11. The red line at 8 indicates the corresponding standard array values. Moving right, we see the distribution for the next highest stat in the array. This continues until the bottom-right, showing the possible outcomes of the highest stat in the array.

The standard array more or less selects the average from rolling, but not always! Interestingly, the largest deviation is the highest stat — rolling more often than not gives 16 or higher, compared to the standard array’s 15.

Seemingly small differences result in the fact that rolling will on average give higher ability scores than the standard array. By looking at the total ability score of the array, which is sum of the whole array, we get a better idea of how they compare. The figure below shows the bell-shaped curved of the total ability scores from rolling compared to the red line of the standard array.

The standard array’s red line is slightly left-of-center of the potential outcomes from rolling. Compared to the standard array’s 72 total ability score, rolling has an expected value of about 73.5. An extra point or two isn’t earth-shattering, but it is a difference worth noting.

Bad luck protection and the upward bias of rolled stats

Armed with an understanding of how different rules generate differing ability scores for a character (point buy is beyond this post’s scope), we shift our focus to how these rules work in practice.

As mentioned in the intro, many groups rightly prioritize the enjoyment of the players over some arbitrary standard of mathematical fairness by rerolling bad arrays. This biases the total ability score upwards by cutting out extreme lows.

One can imagine that a player who has the misfortune to roll significantly under the standard array may want a reroll. This has the effect of chopping off the left side of the total ability score distribution and increasing the odds of the remaining outcomes.

The chart below shows how that shift happens — the underlying yellow-orange bars represent the full possible outcomes of rolling, whereas the slightly higher overlaid blue bars show how the distribution changes when it gets cuts off. The bad luck threshold shown here is the point where we say “any stats that are worse than this should be rerolled”.

The relatively modest bad luck threshold shown above, where a player rerolls when their total ability score is 60 or below (which are effectively commoner stats!), has the same shape as before, but with the left tail cut off. Although the difference seems minor, it is enough to raise the expected total ability score from 73.5 to 74. By allowing players to reroll commoner-tier stats, they expect to gain half of an ability score in the process.

Every bad luck threshold we pick gives a different expected boost to your ability score.

The figure below shows just how that bonus varies. The horizontal axis represents the different thresholds we can set. Higher bad luck thresholds correspond to being more generous with rerolls. The vertical axis is our expected total ability score when we roll using this bad luck protection rule.

As we increase the threshold, we see our expected stat total increase (as represented by the blue markers). Our previous threshold of 60 puts us right at a full ability score increase above the standard array. Increasing it to a higher value of 68 or so is equivalent to two full ASIs above the standard array.

When phrased in terms of ASIs, it’s clear that rolling RAW already gives a notable average boost over the standard array. Adding even a small amount of bad luck protection to that pushes the disparity even further.

As mentioned previously, a bad luck threshold of 60 — which corresponds to an average ability score of 10 — is enough to push rolling to a full ASI above the standard array. A more generous threshold of 66, or an average ability score of 11, doubles the expected stat bonus of rolling from 1.5 to 3 above the standard array.

Of course, a single number like the total ability score cannot capture all of the nuances that come into play when determining what is a “good” stat array. Having a very strong primary stat can more than make up for a poor dump stat or two, and none of that nuance is appropriately captured by this post. Regardless, when looking at different methods for generating ability scores, it’s important to explore how they are used in practice for an honest comparison between them.

Some thoughts on fairness

There is nothing wrong with ensuring everyone is happy by allowing players to rerolls stats, but tables should be wary as to how this method can affect players that use different stat generating methods. If some players want to roll and others prefer to use the standard array, bad luck protection can lead to a lopsided power balance in favor of the rollers.

On one hand, it is something that could easily go by unnoticed and (in all honesty) likely wouldn’t have too much of an effect on the game. On the other hand, it could feel bad to effectively give an ASI to some players at character creation and not to others. Encouraging everyone to use the same method for generating ability scores ensures an even playing field, but if a table does split, buffing the standard array by an ASI isn’t a half-bad way to go.

Another point that often comes up in these discussions about rolling for stats is the opposite problem — good luck protection. Players who roll far above their party members may feel guilty about their greater character power and choose to reroll to get a more average array. This has a counteracting effect, pushing the expected ability scores lower. One strategy to constrain extreme ability scores is to reroll on both overly weak and overly strong characters. If the reroll rules are symmetric, it leaves the expected stats unchanged from the original method.

The goal of character creation should always be to create a party of adventurers that your table is excited to play. Keeping in mind that something as simple as allowing a player to reroll their stats will inflate them is worth keeping in mind, so that you can ensure balanced table where everyone can shine.

Group checks depend too much on whether you have an odd or even number of players - here's a proposed fix

Thu, 22 Apr 2021 00:00:00 GMT

In DnD 5e, a group check is a helpful mechanic to approach situations where the entire party succeeds or fails together. While group checks don’t always come up in the average session, they certainly have their place. In my games, I often use a group Stealth check if the party tries to collectively sneak past an enemy or to adjudicate a Survival check to see if they navigate a hazardous terrain effectively.

Unfortunately — perhaps due to the designers assuming they are rare — the rule determining group checks is very sensitive to whether you have an odd or even number of party members.

Let’s take a look at the rule as written:

To make a group ability check, everyone in the group makes the ability check. If at least half the group succeeds, the whole group succeeds. Otherwise, the group fails. — Player’s Handbook p. 175

The rule itself is simple, but issues arise in the: “at least half the group” clause. As we know, party sizes in DnD are typically somewhere between 3-6 players. If you happen to have an odd number of players in your group, you’re stuck with interpreting what that “at least half” means — do you round up or down?

That decision is best left to the rules lawyers. Instead we focus on the fact that either decision leads to a situation where your party is punished or rewarded just because they have an odd number of members.

The below figure shows what I’m talking about. (Here I assume each party member has a 50% chance of success, which is not necessarily realistic but doesn’t change the result).

Every time you hit an odd numbered group size, the success chance shoots up or down compared to an even sized party, with the direction being determined by how you round. These are the sorts of odd results you end up with when you take a simple rule and combine it with having to round small numbers.

A modification of the rule can alleviate this issue. I propose the following change to the rule as written:

To make a group ability check, everyone in the group makes the ability check. If the group has an even number of members, the whole group succeeds if at least half the group succeeds. Otherwise, the group fails.
If the group has an odd number of members, the DM rolls a dice and records if the result is odd or even. If the DM rolls odd, the whole group succeeds if at least half the group succeeds, rounded down. If the DM rolls even, the whole group succeeds if at least half the group succeeds, rounded up. Otherwise the group fails.

Essentially, we change the rule so that with an odd number of players, the DM randomly chooses between rounding up or down the number of players required for success. This randomization process results in a compromise between the two extremes.

The below figure shows how the proposed rule compares to the original:

The success chance of our new rule is shown in red. It is clearly more well behaved when it comes to odd numbered group sizes. I’d go so far to argue that the proposed rule is more fair to the players — no group should be significantly affected by whether they have an odd or even number of members!

The downside of the proposed rule is that it places a little more burden on the DM by requiring an extra roll, but I believe the added complexity is worth it. Maybe I am the only DM who uses group checks and it doesn’t really matter, but hey, hopefully there was something interesting here to think about.

Calculating attack damage in DnD 5e

Mon, 29 Mar 2021 00:00:00 GMT

Calculating damage is a common — perhaps the most common — task in analyzing builds and theorycrafting in Dungeons and Dragons. It is not an easy one, however. Like many other concepts in DnD, damage comes at the whims of dice. For this reason, damage cannot be effective described as a single number, but rather as a distribution of possible values. On top of that, there are a number of rules and spell-specific effects that must be accounted for when determining the range of outcomes.

In this post we’ll lay out the general framework of the damage distribution and calculate it for a basic attack roll.

Describing damage as a probability distribution

When a character attacks or casts a spell there are typically two steps that occur: 1) a roll by either the player or the DM to determine the degree to which the target is affected and 2) a subsequent damage roll that determines how much damage is taken by the target. There is randomness at every step of this process so we can best represent its outcome as a probability distribution $p(d)$ , which is the probability that an attack does $d$ damage.

Let’s think about how we can compute this for a standard attack roll. We start by recognizing that a target can be affected by an attack in three ways: a miss, a normal hit, or a critical hit. A miss deals no damage, a normal hit will deal damage as written in the ability block, and a critical hit deals damage but with double the number of dice rolled. These three possibilities can be represented by breaking down our damage into conditional components:

p(d) = \sum_{h\in H} p(d \,|\, h) \,p(h)

Where $H = \lbrace \textrm{Miss}, \textrm{Normal}, \textrm{Crit} \rbrace$ . The above expression states that the probability that $d$ damage is done is the weighted sum of doing that damage under each of our three hit types. By breaking it down like this, we can directly compute $p(d)$ by first looking at the hit probabilities, $p(h)$ , and conditional damage distributions, $p(d|h)$ .

It’s not too hard to see that this general formula can represent other damage processes. For a spell with a saving throw we could have $H = \lbrace \textrm{Success}, \textrm{Fail} \rbrace$ . From there, it comes down to defining the conditional damage distributions.

The hit probabilities of an attack roll

The first step of computing our attack damage distribution is calculating the hit probabilities $p(h)$ . We will only concern ourselves with a standard roll, no advantage or disadvantage. Their effects could be explored with the appropriate substitution of probabilities given by advantage or disadvantage rules, and following the structure of the rest of this post.

We’ll start with a critical hit. The rules state a critical hit occurs when rolling a 20 on an attack roll. For a standard attack roll, this is simply:

p(\textrm{crit}) = \frac{1}{20}

Next, we examine the probability of a normal hit, i.e. a roll that meets or exceeds a target’s AC after bonuses but is neither a critical hit or a critical fail (which guarantees a miss). To do this, we must reach back to our previous post on dice rolls probabilities.

We introduce two new parameters: a target roll, $t$ , and a bonus to add to the roll, $b$ . In an attack roll, $t$ is typically the target’s armor class. The parameter $b$ is usually a combination of a proficiency bonus and a stat modifier.

As an example, suppose a fighter with a +5 to hit attempts to strike a Goblin with AC 15, then $t = 15$ and $b = 5$ .

If $X$ is the outcome of our attack roll, then we can write:

p(\textrm{normal}) = \begin{cases} P(t-b \leq X \leq 19)& t-b \geq 2\\ P(2 \leq X \leq 19)& \mathrm{otherwise} \end{cases}

Using our dice roll probabilities, we can write this as:

p(\textrm{normal}) = \begin{cases} \frac{19 - t + b + 1}{20}& t-b \geq 2\\ \frac{18}{20}& \mathrm{otherwise} \end{cases}

The above expression describes the probability that a roll meets or exceeds the target after bonuses, while excluding critical hits and failures.

The last category to look at is the probability of a miss. We could calculate it directly as before, but it’s easier to just derive the result using the complement:

p(\textrm{miss}) = 1 - p(\textrm{crit}) - p(\textrm{normal})

Then we can write:

p(\textrm{miss}) = \begin{cases} \frac{t - b - 1}{20}& t-b \geq 2\\ \frac{1}{20}& \mathrm{otherwise} \end{cases}

With this, we fully describe the probability of an attack hitting when accounting for the target’s AC and the attacker’s bonuses. From here, we turn our attention to the conditional damage distributions.

Damage conditional on the attack roll

The damage distributions themselves are determined by the specific attack type or spell being used, but most follow a pattern — the total damage is determined by the sum of rolling one or more dice and adding a bonus. We will write ${}_{m}D_n$ to represent the random variable of a sum of $m$ $n$ -sided dice and $d_0$ will be the damage bonus added, then:

p(d|h) = P({}_{m}D_n + d_0 = d | h) = P({}_{m}D_n = d - d_0| h)

With the way we’ve written the above, we can see that our damage distribution can be written directly in terms of ${}_{m}D_n$ , which we conveniently defined in a prior post.

For our normal damage distribution, we write it out directly:

p(d|\textrm{normal}) = \frac{1}{n^m}\sum_{l=0}^{\left\lfloor\frac{d - d_0 - m}{n}\right\rfloor} (-1)^l {m \choose l} {d - d_0 - nl - 1 \choose m - 1}

Which is the full expression of our dice roll sum distribution equal to $d - d_0$ as above.

Critical hits are a simple modification of the above, substituting $m \to 2m$ , which represents doubling the number of damage dice that one rolls. This gives:

p(d|\textrm{crit}) = \frac{1}{n^{2m}}\sum_{l=0}^{\left\lfloor\frac{d - d_0 - 2m}{n}\right\rfloor} (-1)^l {2m \choose l} {d - d_0 - nl - 1 \choose 2m - 1}

Our last distribution — damage when the attack roll is missed — is the simplest:

p(d|\textrm{miss}) = \begin{cases} 1& d = 0\\ 0& \textrm{otherwise} \end{cases}

When an attack is missed, damage done is 0 with probability 1. Simple enough.

Complete damage distributions

We now have all the necessary components to compute $p(d)$ in its entirety. As a reminder, for an attack roll we have:

p(d) = p(d|\textrm{miss})p(\textrm{miss}) + p(d|\textrm{normal})p(\textrm{normal}) + p(d|\textrm{crit})p(\textrm{crit})

It would be too verbose to substitute all of our values directly into this expression, but easy enough to describe the output by looking at graphs of different damage distributions. Let’s consider some examples:

Suppose Ria, a fighter, is facing down a zombie. Her attack bonus with her longsword is +5 and the zombie’s AC is 8. Ria’s longsword deals 1d8 + 3 slashing damage. What is the damage distribution of her strike?

From the prompt, we know that $t = 8$ , $b=5$ , $m=1$ , $n=8$ , and $d_0 = 3$ . All there is to do now is plug these values into our previously described expressions. The below graph shows the probable outcomes of her strike.

It’s not exactly a bell curve to say the least. The complex behavior is the mixture of our three possible damage conditions — the single bar at 0 is the probability of missing, the mostly uniform damage from 4 to 11 is the damage from a normal hit, and the triangular distribution that lies on top of that is the critical hit damage.

The shapes and relative magnitudes depend on all the parameters present. To see just how different they can be, let’s consider another situation.

An adventuring party is going toe to toe with a large Ogre. Yar, a cleric, calls on the power of his deity to cast Inflict Wounds at 1st level. His spellcasting ability modifier is +5 and the AC of the Ogre is 11. Damage dealt on a hit is 3d10 necrotic damage. What is the damage distribution?

We approach this in a similar way to before, noting that $t = 11$ , $b=5$ , $m=3$ , $n=10$ , and $d_0 = 0$ .

Now we see an entirely different distribution. The 0 damage possibility has risen to 25% and the bulk of the non-zero damage forms a familiar bell curve from the 3-dice sum with a long tail that represents possible crit damage.

In both of these examples, the mixture of different attack roll outcomes gives rise to non-trivial damage outcomes. Even so, the damage from an attack roll that we explored here is one of the simplest damage mechanics in 5e. Varying spell effects, different damage types, and other features all paint a rich mechanical picture. It suffices to say that we still have much to explore in this realm, but that will be for future posts.

How bad is rolling for health in DnD 5e?

Mon, 08 Mar 2021 00:00:00 GMT

Progression in Dungeons and Dragons revolves around gaining levels, in which your character becomes more powerful. One component of leveling up is increasing your maximum hit points each level. The rules as written provide two alternate ways of determining how much of an increase you get:

Each time you gain a level, you gain 1 additional Hit Die. Roll that Hit Die, add your Constitution modifier to the roll, and add the total to your hit point maximum. Alternatively, you can use the fixed value shown in your class entry, which is the average result of the die roll (rounded up). — Player’s Handbook p. 15

An interesting bit here is that the fixed health rule uses the rounded up average of the roll. This means that the fixed value will be better than rolling, on average. In this post, we’ll dig in a bit deeper, talk about the math behind each rule, and compare the practical differences.

Defining the strategies

Our approach to comparing these rules is to define them mathematically as random variables. As for notation, we will use $X_{\mathrm{roll}}^k$ and $X_{\mathrm{fixed}}^k$ , to represent the amount of health gained after leveling up $k$ times for the rolling and fixed health rules, respectively.

The fixed health rule increases your character’s health pool by the same amount each level, which we write as:

X_{\mathrm{fixed}}^k = k\left(\frac{d+2}{2} + C_{mod}\right)

Where $d$ is the value of your class’s hit dice and $C_{mod}$ is your constitution modifier. As an example, suppose you are playing a Bard with a +1 Constitution modifier, then $X_{\mathrm{fixed}}^k = 6k$ . Each level you gain 6 health (until your constitution modifier changes).

The rolling strategy is a bit more complex. If we let $Y_d^k$ represent the sum of $k$ $d$ -sided dice rolls, we can write the health gained from the dice rolling rule as:

X_{\mathrm{roll}}^k = k C_{mod} + Y_d^k

Since we are primarily interested in comparing these two rules, we can look directly at the difference between them, $\Delta X^k_d = X_\mathrm{fixed}^k - X_\mathrm{roll}^k$ , which represents how much better the fixed strategy is compared to rolling after $k$ level ups of a class with hit dice $d$ . We can rewrite it using the definitions above:

\Delta X^k_d = k \frac{d+2}{2} - Y_d^k

Our dependence on the constitution modifier drops completely and we have a random variable defined in terms of the sum of dice rolls.

Expected differences

A simple step from here is to take a look at the expectation of this difference between the rules. Applying the tools that we looked at in a post on expectations of dice rolls we can tackle this:

\begin{aligned} E\left(\Delta X^k_d\right) &= k \frac{d + 2}{2} - E Y_d^k \\ &= k\frac{d+2}{2} - k \frac{d+1}{2} \\ E\left(\Delta X^k_d\right) &= \frac{k}{2} \end{aligned}

Simply put: the fixed health increase will give you 1 more hit point every 2 levels, on average. It’s no surprise that this value comes up, because it is exactly the amount that is gained by rounding up the average roll each level.

The probability distribution of how much better fixed health is

The expectation is not the whole picture, of course. The whole point of rolling is the element of randomness in the process. We can do a bit of algebra to write out the probability distribution of $\Delta X^k_d$ in terms of the sum of dice rolls:

P(\Delta X^k_d = x) = P\left(Y_d^k = k\frac{d+2}{2} - x\right)

Using the results from our recent post on dice roll sums, we can write out the exact form:

P(\Delta X^k_d = x) = \frac{1}{d^k}\sum_{l=0}^{\left\lfloor\frac{k}{2} - \frac{x}{d}\right\rfloor} (-1)^l {k \choose l} {k\frac{d+2}{2} - x - dl - 1 \choose k - 1}

It’s a bit messy, but we can calculate some interesting quantities, such as the probability that the fixed health increase on level up is strictly better than rolling — $P(\Delta X^k_d > 0)$ — by adding up terms when $x > 0$ . The below graph shows how this quantity changes as you level up:

There are a few interesting features here. The first is how the benefits of taking the fixed health increase are greater with a low hit dice — by 5th level a Wizard or Sorcerer has a 2-in-3 chance of having more health by taking fixed increases. On the other hand, a class with a d10 hit dice doesn’t reach that same likelihood until around 10th or 11th level.

Conceptually, this arises because the $\frac{1}{2}$ average benefit is fairly large relative to the range of possible outcomes on a d6, but on a d12? Not so much.

The second interesting point on this graph (to me) is that there is no difference between fixed health and rolling if you only do it once! So if you’re one of those people who wants to mathematically min-max your character to the greatest degree possible, you can still have a little fun by rolling for health just one time.

Just how big is the difference?

While the above is more focused on whether there is any difference at all, probably what would be more interesting is looking at the range of possible outcomes and seeing how the actual hit point increases vary between the different rules.

In the below figure, we compare the fixed health rule (horizontal line) with the 50% central interval of outcomes from the rolling strategy, you can think of the interval as the range of possible outcomes that are pretty close to average.

We are only showing the results for d6 and d12 hit dice up to 11th level to avoid cluttering the graph, but the behavior we see is consistent for the other hit dice as well. The key feature is that the health pool of the fixed increases inches its way to the upper end of the central interval. By 11th level, fixed health increases are either just outside the 50% interval as is the case for d6 hit dice, or at the upper end of the range as we see for d12.

In all cases, the fixed health outcome is relatively close to the central intervals, which means that the difference between the two rules isn’t that big. If I had to sum it up, I’d say that although the odds favor the fixed health strategy, it wouldn’t be surprising if someone happened to roll higher health in the long term.

Which strategy to pick?

The way you choose to advance your character’s health is one of the more minor decisions you have to make in DnD. But hey. That doesn’t stop us from talking about the pros and cons of each rule:

Why choose fixed increases?

Mathematically, it wins on average
Especially when playing a low hit dice class like Wizard, hit points come at a premium and taking the fixed value is a prudent choice
At low levels, rolling a 1 or 2 for health can have fatal consequences — avoiding that by taking fixed increases could be the difference between life and death for your character

Why choose rolling for health?

Rolling is fun!
The benefits of fixed increases for are smaller for high hit dice classes, so you risk less by choosing to roll
Fixed health prevents you from getting low values, but it also prevents you from getting high values as well — if you want to press your luck, there’s no better way to get a big health pool than some lucky rolls

Outside of any number-based arguments, just pick the strategy that is most likely to give you a character/style of play that you have the most fun with. It is a game, after all.

Deriving the probability distribution for the sum of many dice rolls

Sat, 20 Feb 2021 00:00:00 GMT

Rolling a whole handful of dice is one of a tabletop gamer’s simple pleasures. The math that governs the behavior of that dice roll is not so simple. We previously explored the cases of rolling just one or two dice. Later, we made the general observation that as you roll many dice the probability distribution tends to concentrate near the expected value.

In this post, we are going to look directly at the process of rolling a whole bunch of dice and derive the probability distribution that governs its behavior. Fair warning - this post is going to get pretty math heavy!¹

Counting possible dice roll outcomes

Beginners in probability often work problems involving dice, coins, or playing cards — really any sort of game of chance. These problems are typically solved not by appealing to probability distribution or random variables, but by counting outcomes. We will ultimately use the same approach here. As a refresher, consider the following problem.

What is the probability of rolling a total of 3 on a 2d6 roll?
To solve this, we need two numbers — the total number of possible outcomes and the number of outcomes that result in a total of 3.
There are 36 total possible outcomes, 6 for each individual dice, and we take the product for the total number. When counting up the outcomes that result in 3, we find only two, $(2, 1)$ and $(1, 2)$ .
The probability is the number of outcomes with a total of 3 divided by the total number of outcomes, or $\frac{2}{36} \approx 0.06$ . It’s easy to check that this answer is identical to one that we would get appealing to the uniform probability distribution that we used in this post.

The above is not too hard, but you can easily imagine how much of a pain this would be if you’re using 10 dice instead of 2. Oof. The good news is that there is a better way!

Let’s take a look at this polynomial. The purpose will become clear in a bit.

f(x) = x + x^2 + x^3 + x^4 + x^5 + x^6

This is the all ones polynomial without the constant 1. If we square this polynomial and expand the terms, we get:

\begin{aligned} f(x)^2 = x^2 &+ 2 x^3 + 3x^4 + 4x^5 +5x^6 + 6x^7 \\ &+ 5x^8 + 4x^9 + 3x^{10} + 2x^{11} + x^{12} \end{aligned}

Look carefully at the coefficients. You may notice that they correspond exactly to the number of ways to get a given outcome on a 2d6 roll — there’s one way to get a 2, two ways to get a 3 (that’s the one we did above), three ways to get a 4, and so on and so forth.

If we want to get the probability of rolling an outcome like 3, we look at the coefficient of $x^3$ . We take that coefficient and divide by 36 to get the probability - $\frac{2}{36}$ just like before.

This is important, because what we’ve shown here is an alternate way to count the outcomes — and therefore compute the probability — of our dice roll outcome. Instead of directly counting, we find polynomial coefficients. The above example is for two 6 sided dice, but the general approach for $m$ $n$ -sided dice is the same. We take the below polynomial and find it’s coefficients.

g(x) = \frac{1}{n^m}\left(\sum_{i=1}^n x^i\right)^m

For convenience, we divide the whole polynomial by the total number of outcomes, $n^m$ to make the coefficients probabilities instead of counts. Formally, our problem here can be written with random variables as follows:

X_i \sim \textrm{Dice}(n) \qquad Y = \sum_{i=1}^m X_i

Then, $P(Y=y)$ is calculated by finding the coefficient of $x^y$ in the polynomial $g(x)$ above.

Some of you may recognize that we are taking a conceptual approach to solving this problem with probability generating functions. We could have gone straight there, but I find the counting line of reasoning to be a satisfying way to see why this works.

Tricky polynomials and series

Just knowing that we can compute probabilities using polynomial coefficients doesn’t mean that it’s easy. In this case, it is possible, but we are going to have to do some not-so-obvious polynomial manipulation to get where we want to go.

We’ll start by rewriting our original polynomial $f(x)$ :

\sum_{i=1}^n x^i = \frac{x(x^n-1)}{x-1}

This can be directly verified by polynomial division. This transforms $g(x)$ into:

g(x) = \frac{1}{n^m} \frac{x^m(x^n-1)^m}{(x-1)^m}

From here, we go term by term. We start by taking a look at $(x^n - 1)^m$ . We can expand this as a sum with the binomial theorem:

(x^n - 1)^m = \sum_{j=0}^m {m \choose j} x^{nj} (-1)^{m-j}

Multiplying through by $x^m$ we get:

x^m(x^n - 1)^m = \sum_{j=0}^m {m \choose j} x^{nj + m} (-1)^{m-j}

This represents the numerator of $g(x)$ . Let’s now turn our attention to the denominator, $(x-1)^{-m}$ . Again we can rewrite this as a sum, but this time as a binomial series, which generalizes the binomial theorem to complex exponents:

(x-1)^{-m} = (-1)^{-m}(1-x)^{-m} = (-1)^{-m}\sum_{k=0}^\infty {k + m - 1 \choose m - 1} x^k

We can now write $g(x)$ in it’s full glory as the product:

g(x) = \frac{1}{n^m} \left(\sum_{j=0}^m {m \choose j} x^{nj + m} (-1)^{m-j} \right) \left((-1)^{-m}\sum_{k=0}^\infty {k + m - 1 \choose m - 1} x^k \right)

We make a slight simplification by bringing the factor of $(-1)^{-m}$ into the first sum to give (we also rewrite $(-1)^{-j} = (-1)^j$ ):

g(x) = \frac{1}{n^m} \left(\sum_{j=0}^m (-1)^{j}{m \choose j} x^{nj + m} \right) \left(\sum_{k=0}^\infty {k + m - 1 \choose m - 1} x^k \right)

The finite sum can be thought of as an infinite series that takes on values of $0$ s when $j > m$ . With this in mind, we can take the Cauchy product of the two series to get the coefficients, which generally states:

\left(\sum_{i=0}^\infty a_i x^i\right) \left(\sum_{j=0}^\infty b_j x^j\right) = \sum_{k=0}^\infty c_k x^k

Where the coefficients $c_k$ are given by:

c_k = \sum_{l=0}^k a_k b_{k-l}

Those coefficients are the key. This whole business we’re going through is to get those coefficients. Remember — those coefficients will be our probabilities. If we do some careful accounting to collect the coefficients, we are left with the following expression:

g(x) = \frac{1}{n^m}\sum_{k=0}^\infty \left(\sum_{l=0}^{\left\lfloor\frac{k - m}{n}\right\rfloor} (-1)^l {m \choose l} {k - nl - 1 \choose m - 1} \right)x^k

Where $\lfloor a \rfloor$ is the floor function that rounds down to the nearest integer.

The probability distribution

To get back to probabilities, we ignore the power series and only look at the coefficients to write our final expression for the probability mass function:

\boxed{ P(Y=y) = \frac{1}{n^m}\sum_{l=0}^{\left\lfloor\frac{y - m}{n}\right\rfloor} (-1)^l {m \choose l} {y - nl - 1 \choose m - 1} }

Needless to say, it’s not the cleanest of expressions, and I wouldn’t want to have to calculate any of these probabilities by hand. It is correct though! Exactly so.

I have a hard time divining what this distribution looks like just from the closed form. We get a better idea if we look at a graph of the distribution. Below we show the shapes of the distribution for 2, 3, and 4 d6 rolls.

From top to bottom, we see the triangular distribution that characterizes the 2d6 roll smoothing out as we add more dice. The distribution quickly takes on a familiar bell shaped curve with as little as 4 dice!

We know from the central limit theorem that the sum approaches a normal distribution for large $m$ , but we seem to be getting pretty close even with small $m$ (at least in shape)! Exploring this property further and looking at convergence to a normal distribution will have to be a topic for the future. Stay tuned!

The derivation is adapted from Introduction to Mathematical Probability (1937) by Uspensky↩

How much is advantage really worth?

Sat, 13 Feb 2021 00:00:00 GMT

Continuing our previous discussions on the Advantage and Disadvantage mechanic in DnD 5e, we are going to examine a straightforward question — how much is advantage worth?

On the surface, the question is open ended — it requires defining “worth” in some way. A decent interpretation is “some measure of how much advantage increases your chance of success”. In our post on probability rules of thumb for the mechanic, we give general statements on the outcome of the dice rolls, such as “you are more likely to get a 20 than rolling below a 7”. It gives a feel for relative likelihoods, but still leaves us stuck thinking in terms of probabilities, which suffer from being challenging to interpret.

Another popular way to summarize advantage is to ground it in familiar DnD terms as a bonus. Bonuses are numbers added (or subtracted) from dice rolls depending on how good (or bad) a character is at a certain action, e.g. a barbarian will usually have a higher bonus for swinging a greataxe than a wizard and therefore a higher chance to hit.

Framing advantage as a bonus has led to the claim that “advantage gives you about a +5 bonus to your roll” — is this true? People quickly point out that such a rule is an oversimplification, but how wrong is it? Let’s dig into the math and explore!

The math

As per usual, we will work out the solution for $n$ sided dice and substitute in d20s at the end. We have two random variables that we want to look at:

X_{adv} \sim \mathrm{Adv}(n) \qquad X \sim \mathrm{Dice}(n)

The left being rolling with advantage, which we explored here, and the right being a simple $n$ sided dice roll, explored here. We want to show when these two have the same probability of success by adding a bonus, $b$ , to the simple dice roll. “Success” in DnD meaning “rolling at or above a target value”. We can represent the target value as $x$ and write:

P(X_{adv} \geq x) = P(X + b \geq x)

In the above, we are most interested in the bonus, $b$ . That tells us how much advantage is “worth” by drawing a direct equivalence with adding a modifier to a normal roll.

We are going to use results from the previous posts linked above to avoid going through any derivations, specifically the cumulative probability expressions:

P(X_{adv} \leq x) = \left(\frac{x}{n}\right)^2 \qquad P(X \leq x) = \frac{x}{n}

We will rearrange our expression that equates advantage and a bonus to make use of these results, like so:

\begin{aligned} P(X_{adv} \geq x) &= P(X + b \geq x) \\ P(X_{adv} \geq x) &= P(X \geq x - b) \\ 1 - P(X_{adv} \leq x - 1) &= 1 - P(X \leq x - b - 1) \\ P(X_{adv} \leq x - 1) &= P(X \leq x - b - 1) \\ \end{aligned}

Now we plug in to get:

\left(\frac{x-1}{n}\right)^2 = \frac{x-b-1}{n}

Simple algebra allows us to solve for $b$ :

b = x - 1 - \frac{(x-1)^2}{n}

One requirement that I will impose is that $b$ must be a whole number. The outcome of a dice roll can only ever be a whole number, and fractional bonuses will never increase your probability of success. This means that the values of $b$ that we can find from this will only be approximate in most cases, but they will represent the bonus that gets as close as possible to the probability of success when rolling with advantage. We can summarize our result as:

\boxed{ b_{adv} = \left\lfloor x - 1 - \frac{(x-1)^2}{n} \right\rceil }

Where the notation $\lfloor a \rceil$ means rounding to the nearest whole number.

The results

By inspecting $b_{adv}$ , we can get some of the main properties — first, the bonus depends heavily on the target roll. As it is a quadratic expression, the graph is a downward facing parabola. By finding the maximum of $b_{adv}$ , we determine where advantage gives the biggest bonus. We denote the maximum to be $\hat{b}_{adv}$ and find:

\hat{b}_{adv} = \frac{n}{4} \qquad \hat{x} = \frac{n}{2} + 1

Substituting in $n=20$ , we see an important result — the maximum bonus of advantage is +5 and occurs when the target roll is an 11. So there is some truth to the statement that advantage is worth +5, it just happens to be the maximum possible bonus and occurs when you need to roll about average.

The best way to get a complete view of the bonus from advantage is to look at the graph.

We immediately see some important observations that can be made. First, if your target roll is anywhere between 8 and 14, then +5 is a good rule of thumb. For all of those targets, the best approximating bonus is +5. Outside of those target rolls in the center, the equivalent bonus drops off significantly. When trying to hit target rolls of 17 or higher, advantage is worth about +3 or less. At the extreme, advantage only gives you a bonus of about +1 to hit a target of 20.

It can be helpful to think about why advantage has this behavior of being equivalent to a strong +5 bonus when you need to hit around an 11 and dropping off when you need to hit higher. It comes down to what is actually happening when you apply the effects of advantage versus adding a bonus.

When you roll with advantage, you shift the mass of the distribution so that low rolls become rarer and high rolls become more likely, but the possible outcomes stay the same. However when you roll with a bonus, it has the effect of shifting the distribution uniformly — with a +5 bonus, it’s no longer possible to roll below a 6 and rolls up to a 25 can be achieved. This shifting causes bonuses to be a stronger force than advantage when trying to hit targets at the edge of possibility.

Okay…so what is it actually worth?

One thing that’s clear from the above is that the relative value of advantage depends heavily on the target roll. Without knowledge of typical target rolls, it can be hard to give a hard and fast rule. To really dig into this question, exploring the rules of a system to figure out a typical target roll for success would be necessary (perhaps a worthwhile exploration in a future post).

That being said, we can get somewhere with just a little bit of reasoning. If we pretend that we are completely ignorant of common target rolls, then we can approximate the overall $b_{adv}$ with the median value over all target rolls. Doing so we find:

\mathrm{Med}(b_{adv}) = 4

An equivalent bonus of +4 is not much different than the commonly used +5, but it is a more conservative rule of thumb. A +5 represents the highest bonus effect of advantage, whereas +4 represents an average over the whole range of outcomes. The more extreme the target roll, the weaker the bonus gets. In practice, it seems likely the the majority of target rolls would fall somewhere near the middle of the distribution, and therefore the commonly accepted +5 rule of thumb effectively captures the typical behavior of rolling with advantage.

A short note on disadvantage, it is fairly easy to show that all the work above applies to disadvantage as well, you just need to make the bonuses negative — a good exercise for the intrepid reader!

Expectations and variances of dice rolls

Sun, 07 Feb 2021 00:00:00 GMT

Often when rolling a dice, we know what we want — a high roll to defeat the monster or win a wager — unfortunately for us, desire has little impact on the outcome of the roll. In these situations, it’s useful to know what to expect and how variable the outcome will be around that expectation.

In this post, we define expectation and variance mathematically, compute them for dice rolls, and explore some key properties that help us understand the potential outcomes.

Some definitions

Just by their names, we get a decent idea of what these concepts tell us. Expectation (also known as expected value or mean) gives us a single value that summarizes the average outcome, often representing some measure of the center of a probability distribution. Variance quantifies how variable the outcomes are about the average. A low variance implies that most of the outcomes are clustered near the expected value whereas a high variance implies the outcomes are spread out.

We represent the expectation of a discrete random variable $X$ as $E(X)$ and variance as $\mathrm{Var}(X)$ . They can be defined as follows:

E(X) = \sum_{k=1}^n k P(X=k) \qquad \mathrm{Var}(X) = E(X^2) - E(X)^2

Expectation is a sum of outcomes weighted by their probability. The variance is itself defined in terms of expectations. As we primarily care dice rolls here, the sum only goes over the $n$ finite outcomes representing the $n$ faces of the dice (it can be defined more generally as summing over infinite outcomes for other probability distributions).

One important thing to note about variance is that it depends on the squared expectation and the expectation of $X^2$ . While we could calculate the probability distribution of $X^2$ and compute the expectation directly, it is much easier to use the law of the unconscious statistician:

E(g(X)) = \sum_{k=1}^n g(k) P(X=k)

This allows us to compute the expectation of a function of a random variable, $g(X)$ , with the original probability distribution and applying the function, $g$ , to the outcomes, $k$ , in the sum.

Applying these to dice rolls

All we need to calculate these for simple dice rolls is the probability mass function, which we explored in our post on the dice roll distribution:

X \sim \textrm{Dice}(n) \qquad P(X=x) = \begin{cases} \frac{1}{n} & x\in\{1,\cdots,n\} \\ 0 & \mathrm{otherwise} \end{cases}

The direct calculation is straightforward from here:

\begin{aligned} E(X) &= \sum_{k=1}^n \frac{k}{n} \\ &= \frac{1}{n} \sum_{k=1}^n k \\ &= \frac{n(n+1)}{2n} \end{aligned}

Yielding the simplified expression for the expectation:

\boxed{ E(X) = \frac{n+1}{2} }

The expected value of a dice roll is half of the number of faces plus $1/2$ . Let’s take a look at the variance — we first calculate $E(X^2)$ :

\begin{aligned} E(X^2) &= \frac{1}{n} \sum_{k=1}^n k^2 \\ &= \frac{1}{n}\frac{n(n+1)(2n+1)}{6}\\ &= \frac{(n+1)(2n+1)}{6} \end{aligned}

Substituting this result and the square of our expectation into the definition for variance we get:

\mathrm{Var}(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2

Simplifying this expression yields:

\boxed{ \mathrm{Var}(X) = \frac{n^2 - 1}{12} }

What does this tell us about dice rolls?

This is the part where I tell you that expectations and variances are mostly useless summaries of single dice rolls. The fact that every face is equiprobable in a single roll is all the information you need to understand the behavior of one dice.

On the other hand, expectations and variances are extremely useful when rolling multiple dice. At the end of our post on simple dice roll probabilities, we showed that when you sum multiple dice rolls, the distribution concentrates about the center of possible outcomes — in fact, it concentrates exactly around the expectation of the sum. We see this for two d6s here:

As we add more dice, the distributions concentrates to the expected value as it approaches a normal distribution.

Since our multiple dice rolls are independent of each other, calculating the expectation and variance can be done using the following true statements (the statement on expectations is always true, the statement on variance is true only if the random variables are uncorrelated):

E\left(\sum_{i=1}^m X_i\right) = \sum_{i=1}^m E(X_i)

\mathrm{Var}\left(\sum_{i=1}^m X_i\right) = \sum_{i=1}^m \mathrm{Var}(X_i)

The expectation and variance of a sum of $m$ dice is the sum of their respective expectations and variances. If we plug in what we derived above, we get expressions for the expectation and variance of a sum of $m$ identical dice:

E\left(\sum_{i=1}^m X_i\right) = \frac{m(n+1)}{2}

\mathrm{Var}\left(\sum_{i=1}^m X_i\right) = \frac{m(n^2 - 1)}{12}

A quick check using $m=2$ and $n=6$ gives an expected value of $7$ , which matches up exactly with the peak in the above graph.

Both expectation and variance grow with linearly with the number of dice. As we said before, variance is a measure of the spread of a distribution, but we can also look at the standard deviation for a more interpretable way of quantifying spread — it is defined as the square root of the variance:

\sigma_X = \sqrt{\mathrm{Var}(X)}

$\sigma_X$ is considered more interpretable because it has the same units as the expected value, whereas variance is measured in terms of squared units (a consequence of all those powers of two in the definition.) On the other hand, standard deviation allows us to use quantities like $E(X) \pm \sigma_X$ to directly summarize the spread of outcomes.

The important conclusion from this is: when measuring with the same units, expectation grows faster than the spread of the distribution, as:

E(\textstyle\sum\! X) \propto m \qquad \sigma_X \propto \sqrt{m}

The range of possible outcomes also grows linearly with $m$ , so as you roll more and more dice, the likely outcomes are more concentrated about the expected value relative to the range of all possible outcomes. This can be seen intuitively by recognizing that if you are rolling 10 6-sided dice, it is unlikely that you would get all 1s or all 6s, and more likely to get a mixture of values which have a tendency to average out near the expected value.

While we have not discussed exact probabilities or just how many of the possible outcomes lie close to the expectation, the main takeaway is the same — when rolling multiple dice, the expected value gives a good estimate for about where you should expect the outcome to be. The more dice you roll, the more confident you should be that the sum will be close to the expectation.

Some helpful rules of thumb about Advantage and Disadvantage

Mon, 01 Feb 2021 00:00:00 GMT

In our previous post, we dug into the mathematics of DnD 5e’s advantage and disadvantage mechanic. Looking at the probability expressions we derived allowed us to understand their general behavior. Here we go through a few helpful rules of thumb to provide more practical results to apply directly in your games.

A few tips

As advantage and disadvantage only applies to d20 rolls in DnD, these results come from our probability expressions with $n=20$ .

Advantage

The single most likely outcome is a natural 20, occurring in about 1-in-10 rolls.
The least likely outcome is a natural 1, occurring in 1-in-400 rolls.
There is about a 1-in-2 chance of rolling a 15 or higher.
There is about a 9-in-10 chance of rolling a 7 or higher.
Rolling a natural 20 is more likely than rolling below a 7.

Disadvantage

The single most likely outcome is a natural 1, occurring in about 1-in-10 rolls.
The least likely outcome is a natural 20, occurring in 1-in-400 rolls.
There is about a 1-in-2 chance of rolling a 6 or lower.
There is about a 9-in-10 chance of rolling a 14 or lower.
Rolling a natural 1 is more likely than rolling above a 14.

Why are these true statements?

A fairly straightforward way to show that all of the above statements are true is to examine the PMF and CDF graphs that we made for d20 rolls at the end of the last post. We show them again below:

By mousing over or tapping on the graphs you can see the exact probabilities of different situations. Mousing over the first graph shows that the probability of a natural 20 with advantage is 0.0975, which is about 1-in-10 and is higher than the probability of rolling a 6 or lower with advantage as seen in the second graph (which is only 0.09). By playing around with these, you can come up with all sorts of useful results!

Here’s a concrete DnD 5e example — a Champion Fighter scores a critical hit when rolling a 19 or higher from its Improved Critical feature. Looking at the graph above, we see that our fighter with advantage will crit on average in 1 out of 5 hits! With an extra attack and Action Surge at their disposal, an advantaged fighter is force to reckoned with.

A derivation of the Advantage/Disadvantage probability distributions

Mon, 25 Jan 2021 00:00:00 GMT

In Dungeons and Dragons 5th edition, the Advantage and Disadvantage mechanic provides a straightforward way to tip the scales of fate in favor (or against) the characters depending on the circumstances they face. The mechanic works like this — when rolling a d20 (20-sided dice) to determine the outcome of an action, you instead roll two d20s and keep either the higher result (with advantage) or the lower (with disadvantage). As a result, the odds of particular outcomes change. In this post, we derive the probability distributions that describe those outcomes to better understand how rolling with advantage or disadvantage works.

Two dice rolls and the comparisons between them

Every dice roll serves as a means to sample from an underlying probability distribution. As discussed in a previous post, this underlying distribution is one that picks a random whole number between 1 and $n$ , the number of sides on the dice, which we can call the dice roll distribution.

Rolling two dice does nothing more than generate a sample of size two instead of one from the dice roll distribution. As with most dice rolls, we assume that the outcome on one dice has no effect on the other, or independence.

We write the following to represent our random sample:

X_1, X_2 \sim \textrm{Dice}(n)

Meaning that $X_1$ and $X_2$ each represent the random outcome of an $n$ -sided dice roll. Although the advantage/disadvantage mechanic typically applies to $n=20$ , it’s straightforward to explore the result for a general value $n$ and substitute in 20 later.

Using our above notation, we mathematically express advantage and disadvantage as the maximum and minimum of the random variables above:

Y_{adv} = \max(X_1, X_2) \qquad Y_{dis} = \min(X_1, X_2)

There’s nothing particularly profound here. We’ve just defined explicitly what advantage and disadvantage are.

A little bit of algebra to get us going

Ultimately, we are interested in the probability that a roll with advantage or disadvantage gives a specific value, e.g. $P(Y_{adv} = y)$ . First, let’s focus on advantage. We can start by writing the following:

P(Y_{adv} = y) = P(\max(X_1, X_2) = y)

The probability that the maximum of two random variables is equal to $y$ can be rewritten as the probability that one of the two variables is equal to $y$ and the other is less than $y$ . Since either $X_1$ or $X_2$ can be the maximum, we just add together the same situation but swapped. There is one edge case this doesn’t capture — if $X_1 = X_2 = y$ — we can add that on to get the full, correct expression:

\begin{aligned} P(\max(X_1, X_2) = y) = P(X_1 = y, X_2 < y) &+ P(X_1 < y, X_2 = y)\\ &+ P(X_1 = y, X_2 = y) \end{aligned}

As $X_1$ and $X_2$ are independent, the joint probabilities become products, and since $X_1$ and $X_2$ come from the same distribution, we can drop the subscript and just call them $X$ to refer to underlying dice roll distribution.

\begin{aligned} P(\max(X_1, X_2) = y) = P(X = y) P(X < y) &+ P(X < y)P(X = y) \\ &+ P(X=y)^2 \end{aligned}

The first two terms on the right side are identical, we combine them to get:

P(\max(X_1, X_2) = y) = 2 P(X = y) P(X < y) + P(X=y)^2

One minor simplification, as $y$ is discrete, we can write $P(X < y) = P(X \leq y-1)$ . This gives (and also using the $Y_{adv}$ notation again)

P(Y_{adv} = y) = 2 P(X = y) P(X \leq y-1) + P(X=y)^2

This gives us the probability mass function of rolling with advantage in terms of the PMF and CDF of the dice roll distribution, $X$ . Before we substitute in those functions, let’s quickly go through a similar process for the disadvantage distribution:

\begin{aligned} P(Y_{dis} = y) &= P(\min(X_1, X_2) = y) \\ &= P(X_1 = y, X_2 > y) + P(X_1 > y, X_2 = y) + P(X_1=y, X_2 = y) \\ \end{aligned}

Which gives, in similar form to above:

P(Y_{dis} = y) = 2 P(X=y)P(X > y) + P(X = y)^2

The probability mass functions

To take our expressions to their simplified forms, we substitute in the functional forms of these probability statements that we found previously. First, we do so for advantage:

P(Y_{adv} = y) = 2\left(\frac{1}{n}\right)\left(\frac{y-1}{n}\right) + \left(\frac{1}{n}\right)^2

Which simplifies to:

\boxed{ P(Y_{adv} = y) = \frac{2y - 1}{n^2} \qquad y\in\{1,\cdots,n\}}

For disadvantage:

P(Y_{dis} = y) = 2\left(\frac{1}{n}\right)\left(1 - \frac{y}{n}\right) + \left(\frac{1}{n}\right)^2

Which simplifies to:

\boxed{ P(Y_{dis} = y) = \frac{2(n - y + 1) - 1}{n^2} \qquad y\in\{1,\cdots,n\}}

The above forms show the clear symmetries between the two processes. A great way to really understand what these probability distributions look like is to plot them, like below:

A few things are immediately obvious — advantage shifts the odds in favor of high numbers, disadvantage does the same for low numbers, and they are mirror images of each other. The symmetry is not terribly surprising given the underlying process is symmetric, but it’s nice to see clearly.

Another point worth making is the single most likely outcome for rolling with advantage is the max value of the dice. For disadvantage it’s rolling a 1. While still unlikely in general, it’s notable that you would expect to see the extreme outcomes more often than any other single value.

The cumulative distribution functions

We can take our work one step further and find the cumulative distribution functions of advantage and disadvantage. Our approach will be to directly calculate:

P(Y_{adv} \leq y) = \sum_{k=1}^y P(Y_{adv} = k) \qquad P(Y_{dis} \leq y) = \sum_{k=1}^y P(Y_{dis} = k)

We’ll start with advantage.

\begin{aligned} P(Y_{adv} \leq y) &= \sum_{k=1}^y \frac{2k - 1}{n^2}\\ &= \frac{2}{n^2}\left(\sum_{k=1}^y k\right) - \frac{1}{n^2} \left(\sum_{k=1}^y 1 \right) \end{aligned}

Using the fact that $\sum_{k=1}^y k = y(y+1)/2$ , the above expression simplifies to:

\boxed{ P(Y_{adv} \leq y) = \left(\frac{y}{n}\right)^2 \qquad y\in\{1,\cdots,n\} }

We will carry out a nearly identical process for disadvantage:

\begin{aligned} P(Y_{dis} \leq y) &= \sum_{k=1}^y \frac{2(n - k + 1) - 1}{n^2}\\ &= \frac{2}{n}\left(\sum_{k=1}^y 1\right) - \frac{2}{n^2} \left(\sum_{k=1}^y k\right) + \frac{1}{n^2} \left(\sum_{k=1}^y 1\right) \end{aligned}

Similar to before we can evaluate these sums and get our final expression:

\boxed{ P(Y_{dis} \leq y) = \frac{2y}{n} - \left(\frac{y}{n}\right)^2 \qquad y\in\{1,\cdots,n\} }

Taking a look at these CDFs graphically shows the differing behavior clearly:

They again appear similar but mirrored — advantage grows slowly at first but picks up as it gets to the higher numbers, disadvantage grows quickly and tapers off as you reach the high numbers.

The median provides a good rule of them for these distributions. When rolling with advantage, you are more likely to roll a 15 or higher than below that. With disadvantage, you are more likely to roll a 6 or below than above that. The fact that the bulk of the distributions are pushed closer to the extremes shows just how much of an effect this mechanic has on the potential outcomes.

Advantage and disadvantage is rather simple to implement but does an excellent job at biasing the outcome of dice rolls in whichever way best suits the situation at hand. Since it’s rise in popularity in DnD 5e, it has proliferated to other tabletop games, and some of the results in this post can go a long way towards getting a strong understanding of how advantage and disadvantage really work.

Probabilities of simple dice rolls

Tue, 19 Jan 2021 00:00:00 GMT

I want to take some time to talk about dice. We all know those small, typically symmetric, solid objects that we toss during gaming and gambling. For millennia, humanity has allowed fate to generate random outcomes by the throw of a dice. As popular as ever, it is worth taking a closer look at how dice behave by exploring the mathematics that describe them.

A single roll

Let’s start by thinking about the most familiar of situations, rolling a single 6-sided dice, with either numbers or pips indicating the value of a face. There’s not too much to say here, if the dice is fair, then each side shows up with probability $\frac{1}{6}$ . So, what’s the probability of getting a 6? 1-in-6. How about the probability of getting a 5 or 6? We just add together the two probabilities to find $\frac{1}{6} + \frac{1}{6} = \frac{2}{6}$ , or 1-in-3. Pretty simple.

While the 6-sided dice (or d6) is the most common, plenty of other dice appear in other contexts, such as the Dungeons and Dragons linchpin: the 20-sided dice (or d20). Instead of going down the list of possible numbers of sides for a dice, let’s just take the general case and say we have an $n$ -sided dice. As for the 6-sided dice, the probability of any one side showing after a roll is $\frac{1}{n}$ . We can formalize our notion of dice roll outcomes a bit by letting $X$ represent the outcome of a roll as a random variable. If our dice is $n$ -sided, then $X$ can take on values $1, 2, \cdots, n$ , each with probability $\frac{1}{n}$ . We can summarize this by writing:

\boxed{ X \sim \textrm{Dice}(n) \qquad P(X=x) = \begin{cases} \frac{1}{n} & x\in\{1,\cdots,n\} \\ 0 & \mathrm{otherwise} \end{cases} }

The notation on the left says that $X$ is distributed according to the dice roll distribution with $n$ sides (we could call it the discrete uniform distribution with endpoints 1 and $n$ , but I think “dice roll” more clearly describes what it is). The right-hand side defines the probability mass function (PMF) of the distribution — giving the probability that a specific face $x$ appears when rolling. Explicitly stating that the probability is zero outside of bounds of the dice is not really necessary, but gives a well-defined answer to questions like: “what is the probability of rolling a 7 on a 6-sided dice?” (it’s zero).

Working with the PMF is great, because any conclusions we draw from using it will be true for dice of any number of sides, we just substitute in a value for $n$ . One simple and useful result is to calculate $P(X \leq x)$ , the probability that a dice roll is less than or equal to a given value, $x$ . This follows pretty simply from the PMF:

\begin{aligned} P(X\leq x) &= P(X=1) + \cdots + P(X=x) \\ &= \sum_{k=1}^x P(X=k) \\ &= \sum_{k=1}^x \frac{1}{n} \\ P(X\leq x)&= \frac{x}{n} \end{aligned}

This expression is the cumulative distribution function. It’s simple to use it to compute the complement, the probability that a dice roll is above some value, $P(X > x)$ :

P(X > x) = 1 - P(X \leq x) = 1 - \frac{x}{n}

Here’s a quick example familiar to anyone who has played a fantasy tabletop RPG:

Alice’s character Ria is facing down a vicious gnoll. Ria strikes out with her longsword. In order to strike true, Alice needs to roll above an 11 on a d20. What is the probability that Ria successfully attacks the gnoll?

Using the above result: $P(X > 11) = 1 - \frac{11}{20} = \frac{9}{20} = 45\%$ . Despite Ria’s skill with a blade, her chances are slightly worse than the flip of a coin.

To try to hammer home what we mean by the probabilities of a single dice roll, the below chart shows the probability of a particular outcome for a 6-sided dice roll — unsurprisingly, it’s flat. You can click and shift-click the bars to calculate the sum of outcomes and get the same results that the CDF would give you:

One roll more

All right, so one dice roll is pretty straightforward. There is not much more to say than what we have above. Most games that use dice often have variety in both the number and types of dice to add complexity and nuance. As an obvious step from the previous section, we will look at what happens when rolling two dice instead of one — such a setup forms the basis of casino games like craps.

The first point to note is that when we roll two dice, the outcome of one has no effect on the outcome of the other — in the parlance of probability, we say that the two dice are independent. Independence tells us that the probability of a joint outcome is the product of the two individual outcomes. If $X_1, X_2 \sim \textrm{Dice}(n)$ represent the two dice rolls, then:

P(X_1 = x_1 \textrm{ and } X_2 = x_2) = P(X_1 = x_1)P(X_2 = x_2)

This allows us to directly calculate particular joint outcomes using the PMF that we define for the single dice roll case. As a quick example:

What is the probability of rolling snake eyes (two 1s) on two 6-sided dice?
Directly from above, we calculate $P(X_1 = 1 \textrm{ and } X_2 = 1) = \frac{1}{6}\cdot\frac{1}{6} = \frac{1}{36} \approx 2.7\%$

While this gives us all the machinery we need to calculate joint outcomes, it is not what we are usually interested in when we roll two dice. More often, we want to know the total value shown. We can use $Y = X_1 + X_2$ to represent the sum of two dice rolls as a random variable. There are a number of ways to calculate the probabilities $P(Y=y)$ — often you see counting up the number of ways to get a particular sum and dividing that by the total possible outcomes. Here we are going to do a little random variable algebra and arrive at the same result:

\begin{aligned} P(Y = y) &= P(X_1 + X_2 = y) \\ &= P(X_1 = y - X_2) \\ &= \sum_{x_2=1}^n P(X_1 = y - x_2 \vert X_2 = x_2)P(X_2=x_2) \end{aligned}

The last step uses the law of total probability to allow us to write this sum as a product of the individual probabilities of $X_1$ and $X_2$ . From before, we know the probabilities are both $\frac{1}{n}$ , but only when their argument is within the range $1,\cdots,n$ . In the above summation, we need to take care to determine when a term is non-zero. We will rewrite our PMF with helpful notation as:

P(X=x) = \frac{1}{n}\mathbf{1}_{\{1,\cdots,n\}}(x)

The boldface, $\mathbf{1}$ , is the indicator function that is 1 when the argument $x$ is within $\{1,\cdots, n\}$ and 0 otherwise. This is just a way to rewrite our piecewise function from before in a more compact way. If we apply this to our expression for $P(Y=y)$ above, we get:

P(Y=y) = \sum_{x_2=1}^n \frac{1}{n^2} \mathbf{1}_{\{1,\cdots,n\}}(y-x_2)\mathbf{1}_{\{1,\cdots,n\}}(x_2)

We can pull the factor of $\frac{1}{n^2}$ out of the sum and the second indicator function is always 1 since the values of $x_2$ that we are summing up are exactly the values for which the PMF is nonzero, so this gives:

P(Y=y) = \frac{1}{n^2} \sum_{x_2=1}^n \mathbf{1}_{\{1,\cdots,n\}}(y-x_2)

Since the indicator function either takes on the value of 1 or 0, we just need to count up the number of times that it takes on a value of 1. We do this by first writing the inequality that would satisfy the indicator function:

1 \leq y - x_2 \leq n \implies y-n \leq x_2 \leq y - 1

Since $x_2$ is bounded by $1$ and $n$ in the summation, we can add those additional bounds to get:

\max(y-n,1) \leq x_2 \leq \min(y - 1 , n)

Now we just count up the number of integers in these bounds, multiply by the factor of $\frac{1}{n^2}$ that we pulled out and we have out PDF for $Y$ :

\boxed{ P(Y=y) = \begin{cases} \frac{\min(y-1,n) - \max(y-n, 1) + 1}{n^2} &y \in \{2,3,\cdots, 2n\} \\ 0 &\textrm{otherwise} \end{cases} }

A simple example shows how we can use it:

Sam is shooting craps in Vegas and rolls two 6-sided dice. He needs the dice to add up to 7 to win his bet. What are the chances of Sam winning?
Directly from the above PMF: $P(Y=7) = \frac{6 - 1 + 1}{36} = \frac{6}{36} = \frac{1}{6}$ , we see Sam will win this bet 1-in-6 times.

A better way to get an idea of how the sum of two dice rolls behaves is to look at the shape of the distribution, shown below. The uniform distribution that characterizes a single roll has been replaced by a symmetric, triangular distribution.

One of the key properties of adding together multiple dice rolls is that the distribution gets narrower and the majority of the results fall near the average outcome. Sums of dice rolls rather quickly approach a normal distribution as you increase the number of dice.

There will be time to explore the world beyond rolling just two dice, but that is another post for another day.